Solve this problem
Hi, Todd, hope this question qualifies, as I have no clue how to solve it
I've provided you a drawing to the right to assist you in coming up with the answers. You need the following information to solve for the Northerly (R1) and Westerly (R2) ....
Point Load (P) = 167.4571
Length (L) between (R1) and (R2) is 1000
Length (b) between (P) and (R2) is 319.442
And that's all you should need to know to solve for (R1) and (R2).
I've provided you a drawing to the right to assist you in coming up with the answers. You need the following information to solve for the Northerly (R1) and Westerly (R2) ....
Point Load (P) = 167.4571
Length (L) between (R1) and (R2) is 1000
Length (b) between (P) and (R2) is 319.442
And that's all you should need to know to solve for (R1) and (R2).
Replies
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Kaustubh KatdareRe: Mechanical engineering: Solve this problem
@Geonana: Todd runs a funny video blog. I'm moving your question to appropriate section. -
CIVILPRINCESS
I think i have the solution to your problem...geonanaHi, Todd, hope this question qualifies, as I have no clue how to solve it
I've provided you a drawing to the right to assist you in coming up with the answers. You need the following information to solve for the Northerly (R1) and Westerly (R2) ....
Point Load (P) = 167.4571
Length (L) between (R1) and (R2) is 1000
Length (b) between (P) and (R2) is 319.442
And that's all you should need to know to solve for (R1) and (R2).
algebric sum of vertical forces=0
i.e. R1+R2=167.4571----(1)
The moment about any point in the beam is =0
lets take the moment about the point where R2 acts
R1x1000-167.4571x319.442=0
=>R1=53.4928
And substitute this in eq.(1)
R1=53.4928
R2=113.9643
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geonanaOk, something is not quite right because the solution needs to be close to these coords: (within 3 km)N 53° 29.000 W 113° 58.000
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heba66Hi geonana ,
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Bye -
CIVILPRINCESS
But why is the solution in degrees.😕geonanaOk, something is not quite right because the solution needs to be close to these coords: (within 3 km)N 53° 29.000 W 113° 58.000
if you apply degree to my solution you are getting the answer..
i.e.When you convert to degree 53.4928=>53degree29minutes
if you know the reason please let me know....:smile: -
geonanaTHX, just needed to convert it...it's going to work! THX again 😀😁
-
Guttu
that person is geocaching puzzler. It's part of a puzzle.CIVILPRINCESSBut why is the solution in degrees.😕
if you apply degree to my solution you are getting the answer..
i.e.When you convert to degree 53.4928=>53degree29minutes
if you know the reason please let me know....:smile: -
mannucivil
this is determinate beam ,solv it by using equilibrium conditiongeonanaHi, Todd, hope this question qualifies, as I have no clue how to solve it
I've provided you a drawing to the right to assist you in coming up with the answers. You need the following information to solve for the Northerly (R1) and Westerly (R2) ....
Point Load (P) = 167.4571
Length (L) between (R1) and (R2) is 1000
Length (b) between (P) and (R2) is 319.442
And that's all you should need to know to solve for (R1) and (R2). -
shubhrajeet tiwalet us apply newton equilibrium eq we have
1)net force in y dir=0
therefore
r1+r2=p---(1)
net moment about r1 =0
i.e
r2*1000=p*(1000-319.442)---(2)
solve simultaneously get ans
how ever to find difflection
generate an poerator=0;{f(x)=0 or (-)eve}
then bending moment eq is: r1*-p* ----(3)
now since
d^sy/dx^2=eq(3)/(e*i);were e is young modulus and i is moment of inertia
integrate twise will give you defliction having to constat of integration
to find there value
put y=0 at point r1 and r2
were y is difliction
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